Teachers - How to help my 3rd/4th graders write and understand this math equation?

[Anonymous]

Anonymous wrote:Simple algebra starts in 1st grade. I have 7 pencils and my friend has some too. Altogether we have 15 pencils. How many pencils does my friend have?

True.

The example the OP gave is not a 3rd grade example.

[Anonymous]

It's from a 3rd gr math workbook. And nobody is depressed because of it. Relax.

[Anonymous]

Algebra does not only mean solving multi-variable equations. Understanding that 7 + 5 = 6 + 6 is the beginning of algebraic learning - the idea that two sides of an equal sign “balance.” Children DO start learning these ideas in 1st grade.

[Anonymous]

Anonymous wrote:in short, the problem says: there're 15 slices of pizza total; there're seven more slices of pepperoni pizza than cheese pizza.

based on that, I wrote these two equations:

P + C = 15, which they (seem to) understand, and
P = 7 + C, which they had a really hard time with

so, how would you teach them to convert "seven more slices of pepperoni pizza than cheese pizza" into "P = 7 + C"? I ended up using a crappy example of 3 = 1 + 2 means 3 is one more than two. but I know they still didn't get it.

BTW, I know this is way beyond their grade levels but the problem is in a 3rd-gr math workbook I bought for my rising 4th grader (I'm OP of that workbook for C student thread) to review his math topics last year. I'm curious how teachers help make this 'transition' happen in their little heads, coz I found it almost impossible to use my languages to convey the idea/reasoning and grasp a new concept, even with my own kids. I really think math is the hardest to teach...

THANKS!

I don't believe that 3rd/4th graders use variables, so you explanation may have confused the kid/. Per a suggestion below, I would make 2 tables listing concrete examples of both conditions.

So, if there are 15 slices of pizza (some cheese and some pepperoni), then list all possible combinations (1 cheese -14 pepperoni; 2 cheese -13 pepperoni; 3 cheese -12 pepperoni; 4 cheese -11 pepperoni; etc.)
Similarly, if there are 7 more pepperoni than cheese, then list some possible combinations (1 cheese and 8 pepperoni; 2 cheese and 9 pepperoni; 3 cheese and 10 pepperoni; 4 cheese and 11 pepperoni; etc.)
Then, circle the parts of the tables that are similar.

[Anonymous]

You could alternatively solve the problem as follows, avoiding algebra.

Start by eating 7 pepperoni slices.

Since 15-7=8, we then have 8 slices left, and the number of cheese pizza slices is now the same as the number of (remaining) pepperoni pizza slices.

Half of 8 is 4, so there are 4 slices of cheese pizza and 4 (remaining) slices of pepperoni.

Thus at the beginning we have 4 cheese pizza slices and 7+4=11 slices of pepperoni pizza.

(Another way is to buy 7 more cheese pizza slices making the number of slices the same between cheese and pepperoni, then proceed similarly).

[Anonymous]

Anonymous wrote:You could alternatively solve the problem as follows, avoiding algebra.

Start by eating 7 pepperoni slices.

Since 15-7=8, we then have 8 slices left, and the number of cheese pizza slices is now the same as the number of (remaining) pepperoni pizza slices.

Half of 8 is 4, so there are 4 slices of cheese pizza and 4 (remaining) slices of pepperoni.

Thus at the beginning we have 4 cheese pizza slices and 7+4=11 slices of pepperoni pizza.

(Another way is to buy 7 more cheese pizza slices making the number of slices the same between cheese and pepperoni, then proceed similarly).

wow, cool solutions, tks a lot! this is quite different but might just work for my kids!

[Anonymous]

Ok, now I need help understanding this. Lol!

Why can't the answer be 8 peperoni slices and 7 cheese slices? Or 9 peperoni slices and 6 cheese slices? It still follows the rules of having 7 more peperoni slices than cheese and equals 15 slices total.

Yes, I understand the algebraic equations in other posts and how they got the answer. I understand why that's the only answer you will come up with using the algebraic equation to solve it, but why is that the only answer? Given the parameters, an answer of 8 peperoni slices and 7 cheese slices follows the rules too, just without using math for the answer.

(Please don't beat up on me for this question. I'm just wondering.)

[Anonymous]

Anonymous wrote:Ok, now I need help understanding this. Lol!

Why can't the answer be 8 peperoni slices and 7 cheese slices? Or 9 peperoni slices and 6 cheese slices? It still follows the rules of having 7 more peperoni slices than cheese and equals 15 slices total.

Yes, I understand the algebraic equations in other posts and how they got the answer. I understand why that's the only answer you will come up with using the algebraic equation to solve it, but why is that the only answer? Given the parameters, an answer of 8 peperoni slices and 7 cheese slices follows the rules too, just without using math for the answer.

(Please don't beat up on me for this question. I'm just wondering.)

Nevermind, I got it. I mis-read 7 more as being "more than 7." Thank goodness DH helps DD with math!

[Anonymous]

Singapore Math - introduces algebraic thinking in a developmentally appropriate way.

Model drawing …

Write the letter C and then draw a rectangle next to it
Underneath it, write the letter P, draw the same sized rectangle as you had for C right underneath the one for C, and then add on 7 smaller boxes at the end
To the right of both rows, draw a bracket and then write the number 15
This shows you that C is a certain amount and P is 7 more than that. It also shows you that C + P = 15.

To an adult with a fully developed brain, this seems silly and too simple. To a 3rd/4th grader whose brain is still developing, this visualization will help if they don't get the equation right off the bat.

[Anonymous]

Many of my 8th grade algebra kids would struggle with this early in the year. They would make a table and guess/check.